Transformation Rules Approximate Patterns and Queries 7.1 Complexity

Theorem 7.1 (4.2) Let T be a set of transformation rules. The transformation testing problem is NP-complete for T with star-free rules, where does not appear on the rhs of atomic rules. Remark: Note that the reenement rule r 0 of Section 2 is not star-free, and thus the above theorem does not apply. Nevertheless, the transformation testing for it is in NP-TIME. (Each application of the rule r shortens the string, thus it can be applied at most jsj times). 2 As before, the problem is decidable for all atomic rules, even when they are not star-free. Theorem 7.2 (4.3) Transformation testing is in NP for all atomic transformation rules. We next reconsider the membership testing problem { testing whether s 2 e for a string s and an approximate expression e. In Section 5.1 we showed that the problem is in NP for star-free rules with bounded shrinkage. This shrinkage refers to the length of strings. To get the same complexity result for the extended language, we further require that the rules have bounded domain shrinkage. Intuitively, bounded domain shrinkage requires that the range of the constants appearing in the output of a rule be smaller than the range of the constants in the input string by at most a constant. Deenition 7.1 For a string s, let max(s) denote the highest constant in s, and min(s) the smallest one. We say that an atomic rule r has bounded domain shrinkage if there is a constant n, such that for all strings s; s 0 ; t, if s 0 2 (s r(t)) then max(s 0) max(s) ? n. and min(s 0) min(s) + n. The following theorem applies to discrete domains. Theorem 7.3 (5.2) The membership testing problem for the extended language over a discrete domain is NP-complete, for star-free transformation rules with length and domain shrinkage bounded by some constants n l and n d resp. Proof: (Sketch) The proof is similar to that of Theorem 5.2, and is based on the fact that all constants involved in the construction of a string s 2 e are from a nite domain, bounded below by min(s) ? jej n 0 d and above by max(s)+jejn 0 d. (Where n 0 d is the maximal domain shrinkage of all the rules in e.) 2 Recall that it is possible to analyze rules syntactically and decide if they have …